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我编写了一些代码来生成三个围绕项目的单独列表。第一个列表是 ID 列表，第二个列表是开始日期列表，第三个列表是结束日期列表。我想将这些列表组合成一个嵌套字典。

item[0]fromstart_date和item[0]from end_date 与item[0]from相关联project_id[0]

project_id = ['project 1','project 2', 'project 3', 'project 4']

start_date = [datetime(2015,1,12), datetime(2015,1,13), datetime(2015,1,11), datetime(2015,1,13)]

end_date = [datetime(2015,1,15), datetime(2015,1,17), datetime(2015,1,15), datetime(2015,1,14)]

期望的输出：

d = {
    'project 1' : {'start date' : datetime(2015,1,12), 'end date' :  datetime(2015,1,15},
    'project 2' : {'start date' : datetime(2015,1,13), 'end date' :  datetime(2015,1,17},
    'project 3' : {'start date' : datetime(2015,1,11), 'end date' :  datetime(2015,1,15},
    'project 4' : {'start date' : datetime(2015,1,13), 'end date' :  datetime(2015,1,14},
}
4

5 回答 5

2

使用 dict 理解和 zip：

d = {a:{"start date":b,"end date":c } for a,b,c  in zip(project_id,start_date,end_date)}

print d

{'project 4': {'end date': datetime.datetime(2015, 1, 14, 0, 0), 'start date': datetime.datetime(2015, 1, 13, 0, 0)}, 'project 2': {'end date': datetime.datetime(2015, 1, 17, 0, 0), 'start date': datetime.datetime(2015, 1, 13, 0, 0)}, 'project 3': {'end date': datetime.datetime(2015, 1, 15, 0, 0), 'start date': datetime.datetime(2015, 1, 11, 0, 0)}, 'project 1': {'end date': datetime.datetime(2015, 1, 15, 0, 0), 'start date': datetime.datetime(2015, 1, 12, 0, 0)}}

如果您想要一个完全有序的字典，请使用：

from collections import OrderedDict
d = OrderedDict()
for a,b,c in zip(project_id,start_date,end_date):
    d.setdefault(a,OrderedDict())
    d[a]["start date"] = b
    d[a]["end date"] = c
print(d)

{'project 1': OrderedDict([('start date', datetime.datetime(2015, 1, 12, 0, 0)), ('end date', datetime.datetime(2015, 1, 15, 0, 0))]),
'project 2': OrderedDict([('start date', datetime.datetime(2015, 1, 13, 0, 0)), ('end date', datetime.datetime(2015, 1, 17, 0, 0))]),
'project 3': OrderedDict([('start date', datetime.datetime(2015, 1, 11, 0, 0)), ('end date', datetime.datetime(2015, 1, 15, 0, 0))]),
'project 4': OrderedDict([('start date', datetime.datetime(2015, 1, 13, 0, 0)), ('end date', datetime.datetime(2015, 1, 14, 0, 0))])}
于 2015-01-22T23:03:58.633 回答
1 
d = {}
for project, start, end in zip(project_id, start_date, end_date):
    d[project] = {'start date': start, 'end date': end}

假设您的所有信息都在列表中排列，这将生成您想要的输出。

于 2015-01-22T23:01:50.513 回答
1

基本上，我们正在遍历项目列表的索引。使用这些索引，我们创建了一个子字典，用于抓取同一索引的其他列表中的元素。然后，我们将此值与项目 ID 键配对。

d = {}

for i in range(len(project_id)):
    sub = {'start date': start_date[i], 'end date': end_date[i]}
    d[project_id[i]] = sub
于 2015-01-22T23:03:18.883 回答
1 
from datetime import datetime

if __name__=='__main__':

project_id = ['project 1','project 2', 'project 3', 'project 4']
start_date = [datetime(2015,1,12), datetime(2015,1,13), datetime(2015,1,11), datetime(2015,1,13)]
end_date = [datetime(2015,1,15), datetime(2015,1,17), datetime(2015,1,15), datetime(2015,1,14)]

d = dict()

for idx in range(0,len(project_id)):
    d[project_id[idx]] = {'start date': start_date[idx], 'end date': end_date[idx]}

print d
于 2015-01-22T23:01:38.683 回答
-1 
from datetime import datetime

d = {a:{"start date":b.strftime("%Y, %m, %d"),"end date":c.strftime("%Y, %m, %d") } for a,b,c  in zip(project_id,start_date,end_date)}

print(d)

结果

{'project 4': {'end date': '2015, 01, 14', 'start date': '2015, 01, 13'}, 'project 2': {'end date': '2015, 01, 17', 'start date': '2015, 01, 13'}, 'project 3': {'end date': '2015, 01, 15', 'start date': '2015, 01, 11'}, 'project 1': {'end date': '2015, 01, 15', 'start date': '2015, 01, 12'}}
于 2015-01-22T23:23:38.900 回答